Restriction Mapping Bioinformatics Algorithms Part 2 František Mráz, KSVI

Bioinformatics Algorithms

Physical Mapping – Restriction Mapping Bioinformatics Algorithms part 2 František Mráz, KSVI

Based on slides from http://bix.ucsd.edu/bioalgorithms/slides.php And other sources

1 Bioinformatics Algorithms Contents

2 Bioinformatics Algorithms Molecular Scissors – Restriction Enzymes • HindII - first restriction enzyme – was discovered accidentally in 1970 while studying how the bacterium Haemophilus influenzae takes up DNA from the virus

• Recognizes and cuts DNA at sequences:

• GTGCAC

• GTTAAC

Molecular Cell Biology, 4th edition 3

Bioinformatics Algorithms Recognition Sites of Restriction Enzymes

Molecular Cell Biology, 4th edition 5

Bioinformatics Algorithms Restriction Maps

• A map showing positions of restriction

sites in a DNA sequence • If DNA sequence is known then

construction of restriction map is a trivial exercise • In early days of molecular biology DNA

sequences were often unknown • Biologists had to solve the problem of

constructing restriction maps without knowing DNA sequences

6 Bioinformatics Algorithms Measuring Length of Restriction Fragments • Restriction enzymes break DNA into restriction fragments. Direction • Gel electrophoresis is a process for of DNA separating DNA by size and measuring movement sizes of restriction fragments • Visualization: autoradiography or fluorescence

7 Bioinformatics Algorithms Physical Map, Restriction Mapping Problem • Definition: Let S be a DNA sequence. A physical map consists of a set M of markers and a function p: M→ N that assigns each marker a position of M in S. N denotes the set of nonnegative integers • For a set X of points on the line, let

δ X = { | x1 - x2| : x1, x2 ∈ X } denote the multiset of all pairwise distances between points in X called partial digest. In the restriction mapping problem, a subset E ⊆δ X (of experimentally obtained fragment lengths) is given and the task is to reconstruct X from E.

8 Bioinformatics Algorithms Full Restriction Digest: Multiple Solutions

• Reconstruct the order of the fragments from the sizes of the fragments {3,5,5,9}

• Alternative ordering of restriction fragments:

• Reconstruction from the full restriction digest is impossible.

9 Bioinformatics Algorithms Three different problems

• One (full) digest is not enough • Use 2 restriction enzymes • Use 1 restriction enzyme, but differently

1. The double digest problem –DDP 2. The partial digest problem –PDP 3. The simplified partial digest problem –SPDP

10 Bioinformatics Algorithms Double Digest Mapping

• Use two restriction enzymes; three full digests: • Δ A – a complete digest of S using A, • ΔB – a complete digest of S using B, and • ΔAB – a complete digest of S using both A and B.

• Computationally, Double Digest problem is more complex than Partial Digest problem

11 Bioinformatics Algorithms Double Digest: Example

12 Bioinformatics Algorithms Double Digest: Example

Without the information about X (i.e. ΔAB ), it is impossible to solve the double digest problem as this diagram illustrates

13 Bioinformatics Algorithms Double Digest Problem

Input: ΔA – fragment lengths from the complete digest with enzyme A. ΔB – fragment lengths from the complete digest with enzyme B. ΔAB – fragment lengths from the complete digest with both A and B.

Output: A – location of the cuts in the restriction map for the enzyme A. B – location of the cuts in the restriction map for the enzyme B.

14 Bioinformatics Algorithms Double Digest: Multiple Solutions

15 Bioinformatics Algorithms Double digest

• The decision problem of the DDP is NP-complete. • All algorithms have problems with more than 10 restriction sites for each enzyme. • A solution may not be unique and the number of solutions grows exponentially. • DDP is a favourite mapping method since the experiments are easy to conduct .

16 Bioinformatics Algorithms DDP is NP-complete

1) DDP is in NP (easy)

2) given a (multi-)set of integers X = {x1, . . . , xn }. The Set Partitioning Problem (SPP) is to determine whether we can

partition X into two subsets X1 and X2 such that

This problem is known to be NP-complete. ∑ = ∑ xx ∈ 1 ∈XxXx2

17 Bioinformatics Algorithms DDP is NP-complete

• Let X be the input of the SPP, assuming that the sum of all elements of X is even. Then set • ΔA = X, ⎧ KK⎫ n • ΔB = ⎨ , ⎬ . with = xK, and ⎩ 2 2 ⎭ ∑ i i =1 • ΔAB = ΔA.

• then there exists an integer n0 and indices {j1, j2,…jn }with n0 n = xx j i ∑∑ j i i =1 ni0+= 1 because of the choice of ΔB and ΔAB. Thus a solution for the SPP exists. Thus SPP is a DDP in which one of the two enzymes produced only two fragments of equal length.

18 Bioinformatics Algorithms Partial Restriction Digest

• The sample of DNA is exposed to the restriction enzyme for only a limited amount of time to prevent it from being cut at all restriction sites.

• This experiment generates the set of all possible restriction fragments between every two (not necessarily consecutive) cuts.

• This set of fragment sizes is used to determine the positions of the restriction sites in the DNA sequence.

19 Bioinformatics Algorithms Multiset of Restriction Fragments

• We assume that multiplicity of a fragment can be detected, i.e., the number of restriction fragments of the same length can be determined (e.g., by observing twice as much fluorescence intensity for a double fragment than for a single fragment) Multiset: {3, 5, 5, 8, 9, 14, 14, 17, 19, 22}

20 Bioinformatics Algorithms Partial Digest Fundamentals

X: the set of n integers representing the location of all cuts in the restriction map, including the start and end

n: the total number of cuts

δX: the multiset of integers representing lengths of each of the fragments produced from a partial digest

21 Bioinformatics Algorithms One More Partial Digest Example X 0 2 4 7 10 0 2 4 7 10 2 2 5 8 4 3 6 7 3 10

Representation of δX = {2, 2, 3, 3, 4, 5, 6, 7, 8, 10} as a two dimensional table, with elements of X = {0, 2, 4, 7, 10} along both the top and left side. The elements at (i, j ) in the table is

xj – xi for 1 ≤ i < j ≤ n.

22 Bioinformatics Algorithms Partial Digest Problem: Formulation

• Goal: Given all pairwise distances between points on a line, reconstruct the positions of those points.

• Input: The multiset of pairwise distances L, containing n (n -1)/2 integers.

• Output: A set X, of n integers, such that δ X = L.

23 Bioinformatics Algorithms Partial Digest: Multiple Solutions • It is not always possible to uniquely reconstruct a set X based only on δX.

• For example, the set X = {0, 2, 5}

and (X + 10) = {10, 12, 15}

both produce δX={2, 3, 5} as their partial digest set.

• The sets {0,1,2,5,7,9,12} and {0,1,5,7,8,10,12} present a less trivial example of non-uniqueness. They both digest into:

{1, 1, 2, 2, 2, 3, 3, 4, 4, 5, 5, 5, 6, 7, 7, 7, 8, 9, 10, 11, 12}

24 Bioinformatics Algorithms Homometric Sets

0 1 2 5 7 9 12 0 1 5 7 8 10 12

1 1 4 6 8 11 1 4 6 7 9 11

2 3 5 7 10 5 2 3 5 7

5 2 4 7 7 1 3 5

7 2 5 8 2 4

9 3 10 2

12 12

25 Bioinformatics Algorithms Partial Digest: Brute Force

1. Find the restriction fragment of maximum length M. M is the length of the DNA sequence. 2. For every possible set

X ={0, x2 , … ,xn-1 , M}

compute the corresponding δX

3. If δX is equal to the experimental partial digest L, then X is the correct restriction map

26 Bioinformatics Algorithms BruteForcePDP

BruteForcePDP(L, n): M ← maximum element in L

for every set of n – 2 integers 0 < x2 < … xn-1 < M

X ← {0,x2,…,xn-1,M}

Form δX from X if δX = L return X output “no solution”

• BruteForcePDP takes O (M n − 2) time since it must examine all possible sets of positions. • One way to improve the algorithm is to limit the values of xi to only those values which occur in L.

27 Bioinformatics Algorithms AnotherBruteForcePDP

AnotherBruteForcePDP(L, n) M ← maximum element in L

for every set of n – 2 integers 0 < x2 < … xn -1 < M from L

X ← { 0,x2,…,xn -1,M }

Form δX from X if δX = L; return X output “no solution” • It is more efficient, but still slow • If L = {2, 998, 1000} (n = 3, M = 1000), BruteForcePDP will be extremely slow, but AnotherBruteForcePDP will be quite fast • Fewer sets are examined, but runtime is still exponential: 2n – 4 O(n )

28 Bioinformatics Algorithms Branch and Bound Algorithm for PDP

1. Begin with X = {0} 2. Remove the largest element in L and place it in X 3. See if the element fits on the right or left side of the restriction map 4. When it fits, find the other lengths it creates and remove those from L 5. Go back to step 2 until L is empty

29 Bioinformatics Algorithms Branch and Bound Algorithm for PDP

WRONG ALGORITHM

30 Bioinformatics Algorithms Defining D(y, X)

• Before describing PartialDigest, first define D(y, X ) as the multiset of all distances between point y and all other points in the set X

D(y, X ) = {|y – x1|, |y – x2|, …, |y – xn |}

for X = {x1, x2, …, xn }

31 Bioinformatics Algorithms PartialDigest Algorithm

• S. Skiena

PartialDigest(L ): width ← Maximum element in L DELETE(width, L) X ← {0, width } PLACE(L, X )

32 Bioinformatics Algorithms

PartialDigest Algorithm (cont’d)

PLACE(L, X ): if L is empty output X return y ← maximum element in L if D(y, X ) ⊆ L Add y to X and remove lengths D(y, X ) from L PLACE(L, X ) Remove y from X and add lengths D(y, X ) to L if D(width – y, X ) ⊆ L Add (width –y) to X and remove lengths D(width – y, X ) from L PLACE(L, X ) Remove (width –y) from X and add lengths D(width – y, X ) to L return analysis

33 Bioinformatics Algorithms An Example

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 } X = { 0 }

analysis

34 Bioinformatics Algorithms An Example

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 } X = { 0 }

Remove 10 from L and insert it into X. We know this must be the length of the DNA sequence because it is the largest fragment.

35 Bioinformatics Algorithms An Example

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 } X = { 0, 10 }

36 Bioinformatics Algorithms An Example

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 } X = { 0, 10 }

Take 8 from L and make y = 2 or 8. But since the two cases are symmetric, we can assume y = 2.

37 Bioinformatics Algorithms An Example

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 } X = { 0, 10 }

We find that the distances from y=2 to other elements in X are D(y, X ) = {8, 2}, so we remove {8, 2} from L and add 2 to X.

38 Bioinformatics Algorithms An Example

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 } X = { 0, 2, 10 }

39 Bioinformatics Algorithms An Example

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 } X = { 0, 2, 10 }

Take 7 from L and make y = 7 or y = 10 – 7 = 3. We will explore y = 7 first, so D(y, X ) = {7, 5, 3}.

40 Bioinformatics Algorithms An Example

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 } X = { 0, 2, 10 }

For y = 7 first, D(y, X ) = {7, 5, 3} = {|7 – 0|, |7 – 2|, |7 – 10|}. Therefore we remove {7, 5 ,3} from L and add 7 to X.

41 Bioinformatics Algorithms An Example

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 } X = { 0, 2, 7, 10 }

42 Bioinformatics Algorithms An Example

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 } X = { 0, 2, 7, 10 }

Take 6 from L and make y = 6. Unfortunately D(y, X ) = {6, 4, 1 ,4}, which is not a subset of L. Therefore we won’t explore this branch.

43 Bioinformatics Algorithms An Example

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 } X = { 0, 2, 7, 10 }

This time make y = 4. D(y, X ) = {4, 2, 3 ,6}, which is a subset of L so we will explore this branch. We remove {4, 2, 3 ,6} from L and add 4 to X.

44 Bioinformatics Algorithms An Example

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 } X = { 0, 2, 4, 7, 10 }

45 Bioinformatics Algorithms An Example

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 } X = { 0, 2, 4, 7, 10 }

L is now empty, so we have a solution, which is X.

46 Bioinformatics Algorithms An Example

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 } X = { 0, 2, 7, 10 }

To find other solutions, we backtrack.

47 Bioinformatics Algorithms An Example

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 } X = { 0, 2, 10 }

More backtrack.

48 Bioinformatics Algorithms An Example

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 } X = { 0, 2, 10 }

This time we will explore y = 3. D(y, X) = {3, 1, 7}, which is not a subset of L, so we won’t explore this branch.

49 Bioinformatics Algorithms An Example

L = { 2, 2, 3, 3, 4, 5, 6, 7, 8, 10 } X = { 0, 10 }

We backtracked back to the root. Therefore we have found all the solutions.

50 Bioinformatics Algorithms Analyzing PartialDigest Algorithm

• Still exponential in worst case, but is very fast on average • Informally, let T(n) be time PartialDigest takes to place n cuts • No branching case: T(n) < T(n-1) + O(n)

• Quadratic • Branching case: T(n) < 2T(n-1) + O(n)

• Exponential

algorithm

51 Bioinformatics Algorithms PDP analysis

• No polynomial time algorithm is known for PDP. In fact, the complexity of PDP is an open problem. • PartialDigest Algorithm by S. Skiena performs well in practice, but may require exponential time. • This approach is not a popular mapping method, as it is difficult to reliably produce all pairwise distances between restriction sites.

52 Bioinformatics Algorithms Simplified partial digest problem

• Given a target sequence S and a single restriction enzyme A. Two different experiments are performed • on two sets of copies of S: • In the short experiment, the time span is chosen so that each copy of the target sequence is cut precisely once by the restriction enzyme. Let Γ = {γ1 , . . . , γ 2N } be the multi-set of all fragment lengths obtained by the short experiment, where N is the number of

restriction sites in S, and • In the long experiment, a complete digest of S by A is performed. Let Λ = {λ1, . . . , λN+1} be the multi-set of all fragment lengths obtained by the long experiment.

53 Bioinformatics Algorithms SPDP

• Example: Given these (unknown) restriction sites (in kb):

• 0 2 8 9 13 16

• We obtain Λ = {2kb, 6kb, 1kb, 4kb, 3kb} from the long experiment.

• The short experiment yields: • 2 14 • 8 8 • 9 7 • 13 3

• Γ = {2kb, 14kb, 8kb, 8kb, 9kb, 7kb, 13kb, 3kb}

54 Bioinformatics Algorithms SPDP

• In the following we assume that Γ = {γ1, . . . , γ2N } is sorted in non-

decreasing order.

• For each pair of fragment lengths γi and γ2N −i+1, we have

γi + γ 2N −i+1 = L, where L is the length of S.

• Each such pair {γi , γ 2N −i+1 } of complementary lengths corresponds to precisely one restriction site in the target sequence S, which is

either at position γi or at position γ2N −i+1.

• Let Pi = 〈 γi , γ 2N −i+1 〉 and P2N −i+1 = 〈 γ2N −i+1 , γ i 〉 denote the two

possible orderings of the pair {γi , γ2N −i+1 } . We call the first component a of any such ordered pair P = 〈 a, b 〉 the prefix of P.

55 Bioinformatics Algorithms SPDP

• We obtain a set X of putative restriction site positions as follows:

For each complementary pair {γi , γ 2N −i+1}, we choose one of the two possible orderings P and P , and then add the i 2N −i+1 corresponding prefix to X.

• Any such ordered choice X = 〈x1, . . . , xN 〉 of putative restriction sites gives rise to a multi-set of integers R = {r , . . . , r }, with 1 N +1

xi if i =1

• ri := xi − xi −1 if i = 2, . . . ,N

{ L − x if i = N + 1. { N

56 Bioinformatics Algorithms SPDP • Simplified Partial Digest Problem (SPDP): Given multi-sets Γ and Λ of fragment lengths, determine a choice of orderings of all complementary fragment lengths in Γ such that the arising set R equals Λ. • Example: In the example above we have • Γ = {2kb, 3kb, 7kb, 8kb, 8kb, 9kb, 13kb, 14kb} • Λ = {2kb,, 6kb, 1kb, 4kb, 3kb}

• We obtain P1 = 〈 2, 14 〉,P 8 = 〈 14, 2 〉,

P2 = 〈 3, 13 〉, P 7 = 〈 13, 3 〉, P = 〈 7, 9 〉, P = 〈 9, 7 〉, { 3 6 P = 〈 8, 8 〉, P = 〈 8, 8 〉. 4 5

Because of the long experiment we obtain Q = {P1, P7, P6, P4} and X = {2, 8, 9, 13}, from which we get R = {2, 6, 1, 4, 3}, our restriction site map.

57 Bioinformatics Algorithms SPDP • Simplified Partial Digest Problem (SPDP): Given multi-sets Γ and Λ of fragment lengths, determine a choice of orderings of all complementary fragment lengths in Γ such that the arising set R equals Λ. • Example: In the example above we have • Γ = {2kb, 3kb, 7kb, 8kb, 8kb, 9kb, 13kb, 14kb} • Λ = {2kb,, 6kb, 1kb, 4kb, 3kb}

• We obtain P1 = 〈 2, 14 〉,P 8 = 〈 14, 2 〉,

P2 = 〈 3, 13 〉, P 7 = 〈 13, 3 〉, P = 〈 7, 9 〉, P = 〈 9, 7 〉, { 3 6 P = 〈 8, 8 〉, P = 〈 8, 8 〉. 4 5

Because of the long experiment we obtain Q = {P1, P7, P6, P4} and X = {2, 8, 9, 13}, from which we get R = {2, 6, 1, 4, 3}, our restriction site map.

58 Bioinformatics Algorithms SPDP – algorithm

• the algorithm generates all possible choices of ordered pairs – when called

with variable i, it considers both alternatives Pi and P2N −i +1.

• During a call, the current list of restriction sites X = 〈 x1, . . . , xk 〉 and the list R = 〈 r , . . . , r , r 〉 of all fragment lengths are passed as a 1 k k+1 parameter. Note that x

• When processing a new corresponding pair of fragment lengths, the last

element rk+1 of the list R is replaced by two new fragment lengths that arise because the last fragment is split by the new restriction site.

• Initially, X and R are empty. Index of the next pair • SPDP (X{ = 〈 x , . . . , x 〉, R = 〈 r , . . . , r , r 〉, i ) 1 k 1 k k+1

Already placed Corresponding The last fragment can be restriction sites fragments split by further restrictions sites

59 Bioinformatics Algorithms

Corresponding fragments The last fragment can SPDP – algorithm Already placed be split by further restriction sites restrictions sites

Algorithm SPDP (X = 〈 x , . . . , x 〉, R = 〈 r , . . . , r , r 〉, i ): Index of the next pair 1 k 1 k k+1 if k = N and R = Λ then print X // output putative restriction sites

else if i ≤ 2N then Consider Pi = 〈 a, b 〉

if b ∉ X then // the reversed ordering of Pi was

// not used

if k = 0 then Set R’ = 〈 a, b 〉, X’ = 〈 a 〉 if a ∈ Λ then call SPDP(X’,R’, i +1)

else

Set p = a − (L − rk+1) and q = L − a // new fragment lengths, // a -(L - r ) equals a – x for k ≥1 k+1 k if p ∈ Λ then { Set R’ = 〈 r , . . . , r , p, q 〉 1 k Set X’ = 〈 x , . . . , x , a 〉 // add a to the set of restriction sites 1 k Call SPDP(X’,R’, i + 1) // continue using a in this tree’s lineage

Call SPDP(X,R, i + 1) // consider other alternative

60 Bioinformatics Algorithms SPDP – algorithm

• Clearly, the worst case running time complexity of this algorithm is exponential. However, it seems to work quite well in practice. • This algorithm is designed for ideal data. In practice there are two problems: 1. Fragment length determination by gels leads to imprecise measurements , down to about 2 − 7% in good experiments. This can be addressed by using interval arithmetic in the above algorithm. 2. The second problem is missing fragments. The SPDP does not suffer from this problem much because both digests are easy to perform. Moreover, the short experiment must give rise to complementary {values and any failure to do so can be detected. The long experiment should give rise to precisely N + 1 fragments.

61 Bioinformatics Algorithms Summary

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